Prove that $A \otimes_{\mathrm{R}} B$ is a projective $\mathbb{Z}$- module. Let $A \in mod-\mathrm{R}$ be projective as a $\mathbb{Z}$ - module and $B \in \mathrm{R} -mod$ be projective as an $\mathrm{R}$-module. Prove that $A \otimes_{\mathrm{R}} B$ is a projective $\mathbb{Z}$- module.
I am trying to consider the projective property of $A$ and $B$ and use the universal property of the tensor product to construct the map that satisfies the projective property for $A \otimes_{\mathrm{R}} B$. 
I take the epimorphism 
$$C \longrightarrow D$$ 
and consider the maps $$\phi : A \longrightarrow C \text{  and } \psi : B \longrightarrow C$$ 
that exist and make the diagram commute according to the definition of projective modules. Unfortunately I cannot construct the bilinear map from $A \times B \mapsto C$ using the maps from $\phi$ and $\psi$ in order to use the universal property of tensor products. 
I am not sure if this is the right path to a solution but any help is appreciated.
 A: For $\Bbb Z$-modules, projective is the same as free.
If $B=R$, then $A\otimes_R R\equiv A$ which by hypothesis is free
over $\Bbb Z$.
If $B$ is free over $R$, then $B$ is a direct sum of copies of $R$,
and then $A\otimes_RB$ is a direct sums of copies of $A$ and again
is free over $\Bbb Z$.
If $B$ is a general projective $R$-module, it's a direct summand
of a free $R$-module $F$, so that $A\otimes_R B$ is a direct
summand of $A\otimes_R F$, which is free over $\Bbb Z$. Thus
$A\otimes_R B$ is also free over $\Bbb Z$.
A: $A \in \mathrm{R}$-mod is projective iff $\mathrm{Hom}_R(A,\cdot )$ is an exact functor. So we only need to prove that $\mathrm{Hom}_\mathbb{Z}(A \otimes_R B,\cdot )$ is exact. But we have that
$$\mathrm{Hom}_\mathbb{Z}(A \otimes_R B,G) \cong \mathrm{Hom}_R(B,\mathrm{Hom}_\mathbb{Z}(A,G))$$ 
Let 
$$0 \longrightarrow X \longrightarrow Y \longrightarrow Z\longrightarrow 0$$
be an exact sequence. Since $A$ is projective as a $\mathbb{Z}$ module we have that:
$$0 \longrightarrow \mathrm{Hom}_\mathbb{Z}(A,X) \longrightarrow \mathrm{Hom}_\mathbb{Z}(A,Y) \longrightarrow \mathrm{Hom}_\mathbb{Z}(A,Z)\longrightarrow 0$$
is an exact sequence. but since $B$ is projective as an $R$ module we have that:
$$0 \longrightarrow \mathrm{Hom}_R(B,\mathrm{Hom}_\mathbb{Z}(A,X)) \longrightarrow \mathrm{Hom}_R(B,\mathrm{Hom}_\mathbb{Z}(A,Y)) \longrightarrow \mathrm{Hom}_R(B,\mathrm{Hom}_\mathbb{Z}(A,Z))\longrightarrow 0$$
But from the isomorpshim we mentioned this is equivalent to:
$$0 \longrightarrow \mathrm{Hom}_\mathbb{Z}(A \otimes_R B,X) \longrightarrow \mathrm{Hom}_\mathbb{Z}(A \otimes_R B,Y) \longrightarrow \mathrm{Hom}_\mathbb{Z}(A \otimes_R B,Z)\longrightarrow 0$$
This means that $\mathrm{Hom}_\mathbb{Z}(A \otimes_R B,\cdot)$ is an exact functor so $A\otimes_R B$ is projective as a $\mathbb{Z}$ module.
